1094 The Largest Generation （25 分）

A family hierarchy is usually presented by a pedigree tree where all the nodes on the same level belong to the same generation. Your task is to find the generation with the largest population.

Input Specification:
Each input file contains one test case. Each case starts with two positive integers N (<100) which is the total number of family members in the tree (and hence assume that all the members are numbered from 01 to N), and M (<N) which is the number of family members who have children. Then M lines follow, each contains the information of a family member in the following format:

ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a family member, K (>0) is the number of his/her children, followed by a sequence of two-digit ID’s of his/her children. For the sake of simplicity, let us fix the root ID to be 01. All the numbers in a line are separated by a space.

Output Specification:
For each test case, print in one line the largest population number and the level of the corresponding generation. It is assumed that such a generation is unique, and the root level is defined to be 1.

Sample Input:

23 13
21 1 23
01 4 03 02 04 05
03 3 06 07 08
06 2 12 13
13 1 21
 

08 2 15 16
02 2 09 10
11 2 19 20
17 1 22
05 1 11
07 1 14
09 1 17
10 1 18

Sample Output:

9 4

解析

#include<algorithm>
#include<cstdio>
#include<iostream>
#include<queue>
#include<string>
#include<cmath>
using namespace std;
const int maxn = 101;
struct node {
	int layer;
	vector<int> child;
	node() :layer(0) { ; }
}Node[maxn];
int LayerTable[101]{ 0 }; 
void BFS() {
	queue<int> Q;
	Q.push(1);
	Node[1].layer = 1;
	while (!Q.empty()) {
		int temp = Q.front();
		Q.pop();
		LayerTable[Node[temp].layer]++;
		for (auto x:Node[temp].child) {
			Q.push(x);
			Node[x].layer = Node[temp].layer + 1;
		}
	}
}
int main()
{
	int N, M;
	scanf("%d %d", &N, &M);
	for (int i = 0; i < M; i++) {
		int father, Num, son;
		scanf("%d %d", &father, &Num);
		for (int j = 0; j < Num; j++) {
			scanf("%d", &son);
			Node[father].child.push_back(son);
		}
	}
	BFS();
	int max = 0;
	for (int i = 1; i < 101; i++) 
		if (LayerTable[max] < LayerTable[i])
			max = i;
	printf("%d %d", LayerTable[max], max);
}